Show $ (2 - b^2c^2) (1 + b^2) \ge (b^2 + c^2)(b + c - 3)^2 $ in some range. Show $ F = (2 - b^2c^2) (1 + b^2) - (b^2 + c^2)(b + c - 3)^2 \ge 0$ in the following two ranges of $(b,c)$:

*

*$b \in [0 \quad 1]$ and $0 \le c\le b$,

*$b \in [1 \quad 1.2]$ and $0 \le c\le 3-2b$.

Equality occurs only at $(b,c)= (1,0)$.
This can be solved by treating $F$ as a polynomial in one of the variables (the other a parameter), and then using Budan's theorem. This has been performed here, but this is a very lengthy business. Does someone have a more direct method? Thanks in advance!
 A: Proof of (1): Clearly, if $b = 0$, the inequality is true. Also, if $c = 0$, the inequality
becomes $(-b^2 + 4b + 2)(b-1)^2 \ge 0$ which is true.
If $b\in (0, 1]$ and $c\in (0, b]$, let $b = \frac{1}{1+s}, c = b \frac{1}{1+t}$ for $s, t \ge 0$.
The inequality becomes
$$\frac{1}{(1+s)^6(1+t)^4}[f(s,t) + 14s^4 - 3s^2 + 2s] \ge 0$$
where $f(s, t)$ is a polynomial with non-negative coefficients. Thus, the inequality is true.
We are done.
$\phantom{2}$
Proof of (2): If $b = 1$, the inequality becomes $c(1-c)(c^2 - 3c + 4) \ge 0$ which is true.
Also, if $c = 0$, the inequality
becomes $(-b^2 + 4b + 2)(b-1)^2 \ge 0$ which is true.
If $b\in (1, 6/5]$ and $c\in (0, 3-2b]$, let $b = 1 + \frac{1}{5}\cdot \frac{1}{1+s}$
and $c = (3-2b)\frac{1}{1+t}$ for $s, t\ge 0$. The inequality becomes
$$\frac{1}{15625(1+s)^6(1+t)^4}g(s, t) \ge 0$$
where $g(s, t)$ is a polynomial with non-negative coefficients. Thus, the inequality is true.
We are done.
A: Not an answer just a remark for the first case:
If we substitute $b=\frac{z}{z+y}$ and $c=\frac{y}{x+y}$ we get this .Remains to plays properly with the three variables or perhaps use Buffalo's way I don't know .
Hope you find something good !
